QFT Commutator for spacelike separation

[PeroK]

Homework Statement

Show that the commutator for field operators is zero for spacelike separations

Relevant Equations

$$\hat \phi(x) = \int \frac{d^3p}{(2\pi)^{3/2}(2E_p)^{1/2}}(\hat{a}_{\vec p}exp(-ip\cdot x)+\hat{a}^{\dagger}_{\vec p}exp(ip\cdot x))$$

I got as far as:
$$[\hat \phi(x), \hat \phi(y) ] = \int \frac{d^3p}{(2\pi)^{3}(2E_p)}(\exp(-ip.(x-y) - \exp(-ip.(y-x))$$
Then I simplified the problem by taking one of the four-vectors to be the origin:
$$[\hat \phi(0), \hat \phi(y) ] = \int \frac{d^3p}{(2\pi)^{3}(2E_p)}(\exp(ip.y) - \exp(-ip.y))$$
We need to show that this is zero for spacelike $y$. But, I can't see how the spacelike condition is relevant. The hint in the book is that we can change $-y$ to $y$ in the second term. If $p$ was a three-vector I can see this, but the zeroth component of $p$ is the positive $E_p$. And, if it's a simple $p$ substitution, then that should work for any $y$, spacelike or otherwise.

Any help would be appreciated.

 

[PeroK]

I might have a solution. Assuming this whole theory is Lorentz invariant, then we could choose coordinates so that $x$ and $y$ are simultaneous. The zeroth term disappears from the inner product and the result follows.

I guess that's the trick.

 

[vanhees71]

Hint: You don't have to assume that the theory is Lorentz invariant, because the integral for sure is, because $\mathrm{d}^3 p/E_p$ is invariant, and $y \cdot p$ is too. That's why the integral is a scalar field operator. Now you can choose $y$ (given that it is space) in the most simple way to evaluate this function. Note that since you only have the four-vector $y$ available it's clear that the expression must be a function of $y^2=y \cdot y$ only!